--- layout: post title: Stress and strains in soil mechanics date: Tue 16 Mar 2021 05:26:21 PM IST categories: ["Academic Writings", "Soil Mechanics and Foundations"] --- -

1 Introduction

We design the substructures to avoid excessive settlement and collapse of the foundation soil. Substructures transfer the load from superstructure to the soil. We assume soil as continua and calculate the stress and strain field by solving the equilibrium equations. This requires constitutive relations for the soils–stress-strain behavior under different conditions.

A mathematical model serves as the constitutive relation for the soil. We design the experiments to understand the soil and propose a mathematical model based on the experimental observations accounting for physics or statistics of the soil behavior. Experimental modeling consists of elemental experiments, mimicking the behavior of representative soil element, and small scale physical experiments, mimicking a boundary value problem. The elemental experiments help in formulating a constitutive model, whereas physical model tests validate their performance with a numerical model.

This text overviews these aspects in the following.

2 Stress

Consider the continua subjected under different boundary conditions. We define the traction at point x⃗\vec{x} on a plane with normal n̂\hat{n} as:

t⃗=t⃗(t,x⃗,n̂)=limδA→0F⃗δA \begin{aligned} \vec{t} &= \vec{t}\left(t, \vec{x}, \hat{n}\right)\\ &= \lim_{\delta{A}\to{0}} \dfrac{\vec{F}}{\delta{A}} \end{aligned}

Where tt is the time, x⃗\vec{x} is the coordinate of the point, and F⃗\vec{F} is the force acting on an infinitesimal area δA\delta{A}. The limδA→0\lim_{\delta{A}\to{0}} changes to limδA→δArve\lim_{\delta{A}\to{\delta{A_{rve}}}} for soils with area of representative volume element as δArve\delta{A_{rve}}.

We implicitly assume the dependence of time and avoid writing it for the sake of clarity.

Figure 1: Cauchy tetrahedron

2.1 Cauchy’s theorem for the existence of stress

Consider an infinitesimal tetrahedron (size δ\delta with vertex at x⃗\vec{x}, fig. 1) and apply balance of linear momentum:

t⃗(n⃗)δA+∑i=13t⃗(−ei⃗)δAi+ρ(b⃗−a⃗)δV=0 \begin{aligned} \vec{t}\left(\vec{n}\right)\delta{A}+\sum_{i=1}^{3}\vec{t}\left(-\vec{e_{i}}\right)\delta{A_{i}}+\rho\left(\vec{b}-\vec{a}\right)\delta{V}=0 \end{aligned}

With infinitesimal tetrahedron, as δ→0\delta\to{0}, limδ→0δVδA=0 \lim_{\delta\to{0}}\dfrac{\delta{V}}{\delta{A}}=0

Using the above: t⃗(n⃗)=−∑i=13t⃗(−ei⃗)(n⃗⋅ei⃗) \begin{aligned} \vec{t}\left(\vec{n}\right) = -\sum_{i=1}^{3}\vec{t}\left(-\vec{e_{i}}\right)\left(\vec{n}\cdot\vec{e_{i}}\right) \end{aligned}

Now, substitute n⃗\vec{n} with ei⃗\vec{e_{i}} in the above:

t⃗(ei⃗)=−t⃗(−ei⃗) \vec{t}\left(\vec{e_{i}}\right) = -\vec{t}\left(-\vec{e_{i}}\right)

Since the choice of ei⃗\vec{e_{i}} is arbitrary:

t⃗(n⃗)=−t⃗(−n⃗) \vec{t}\left(\vec{n}\right) = -\vec{t}\left(-\vec{n}\right)

Which is the Newton’s law of action and reaction.

We can write the traction on plane normal to n⃗\vec{n} as:

t⃗(n⃗)=∑i=13t⃗(ei⃗)(n⃗⋅ei⃗) \vec{t}\left(\vec{n}\right)=\sum_{i=1}^{3}\vec{t}\left(\vec{e_{i}}\right)\left(\vec{n}\cdot\vec{e_{i}}\right)

The Cauchy stress is defined as:

σ‾(x⃗)=∑i=13t⃗(x⃗,ei⃗)⊗ei⃗ \bar{\sigma}\left(\vec{x}\right)=\sum_{i=1}^{3}\vec{t}\left(\vec{x},\vec{e_{i}}\right)\otimes\vec{e_{i}}

The Cauchy stress tensor σ‾(x⃗)\bar{\sigma}(\vec{x}) (at point x⃗\vec{x}) maps the normal n⃗\vec{n} (on a given plane through point x⃗\vec{x}) to traction t⃗(x⃗,n⃗)\vec{t}(\vec{x},\vec{n}) (acts at point x⃗\vec{x}, on plane normal to n⃗\vec{n}):

t⃗(x⃗,n⃗)=σ‾(x⃗)n⃗ \vec{t}\left(\vec{x},\vec{n}\right)=\bar{\sigma}\left(\vec{x}\right)\vec{n}

2.2 Stress components

We often deal with stress components (σij\sigma_{ij}) than the stress tensor. The first index represents the plane and second index represents the direction in which it acts. For example, σ23\sigma_{23} represents component of the stress tensor which acts on plane normal to x2x_{2} along the direction x3x_{3}. If both indices are equal, it represents the normal stress, and different indices represent shear stress.

Figure 2: Stress components

2.3 Balance of linear momentum

Net force acting on a spatial region Ωt\Omega_{t} is equal to its rate of change of linear momentum.

∫∂Ωtt⃗(n⃗)da+∫Ωtρb⃗dv=∫Ωtρv⃗dv‾̇ \begin{aligned} \int_{\partial{\Omega}_{t}}\vec{t}\left(\vec{n}\right)da+\int_{\Omega_{t}}\rho\vec{b}dv=\dot{\bar{\int_{\Omega_{t}}\rho\vec{v}dv}} \end{aligned}

Using the Cauchy’s theorem and Reynolds transport theorem: ∫∂Ωtσ‾n⃗da+∫Ωtρb⃗dv=∫Ωtρv⃗̇dv \begin{aligned} \int_{\partial{\Omega}_{t}}\bar{\sigma}\vec{n}da+\int_{\Omega_{t}}\rho\vec{b}dv=\int_{\Omega_{t}}\rho\dot{\vec{v}}dv \end{aligned}

Using Gauss’s divergence rule: ∫Ωt∇⋅σ‾dv+∫Ωtρb⃗dv=∫Ωtρv⃗̇dv∫Ωt(∇⋅σ‾+ρb⃗−ρv⃗̇)dv=0 \begin{aligned} \int_{\Omega_{t}}\nabla\cdot\bar{\sigma}dv+\int_{\Omega_{t}}\rho\vec{b}dv=\int_{\Omega_{t}}\rho\dot{\vec{v}}dv\\ \int_{\Omega_{t}}\left(\nabla\cdot\bar{\sigma}+\rho\vec{b}-\rho\dot{\vec{v}}\right)dv=0 \end{aligned} The region Ωt\Omega_{t} is arbitrary, this leads to ∇⋅σ‾+ρb⃗−ρv⃗̇=0 \nabla\cdot\bar{\sigma}+\rho\vec{b}-\rho\dot{\vec{v}}=0

2.4 Balance of angular momentum

Net moment of a spatial region Ωt\Omega_{t} is equal to its rate of change of angular momentum.

∫∂Ωtx⃗×t⃗(n⃗)da+∫Ωtx⃗×ρb⃗dv=∫Ωtx⃗×ρv⃗dv‾̇ \begin{aligned} \int_{\partial{\Omega}_{t}}\vec{x}\times\vec{t}\left(\vec{n}\right)da+\int_{\Omega_{t}}\vec{x}\times\rho\vec{b}dv=\dot{\bar{\int_{\Omega_{t}}\vec{x}\times\rho\vec{v}dv}} \end{aligned}

Using the Cauchy’s theorem, balance of linear momentum and Gauss’s divergence rule: ∫∂Ωtx⃗×(σ‾n⃗)da−∫Ωtx⃗×(∇⋅σ‾)dv=0 \begin{aligned} \int_{\partial{\Omega}_{t}}\vec{x}\times\left(\bar{\sigma}\vec{n}\right)da-\int_{\Omega_{t}}\vec{x}\times\left(\nabla\cdot\bar{\sigma}\right)dv=0 \end{aligned}

We have to apt for the indicial notation: ∫Ωtσjiϵijkek=0 \begin{aligned} \int_{\Omega_{t}}\sigma_{ji}\epsilon_{ijk}e_{k}=0 \end{aligned}

Since Ωt\Omega_{t} is arbitrary: σjiϵijk=0(1) \sigma_{ji}\epsilon_{ijk}=0 \qquad(1)

Now we have: σij=σji \sigma_{ij}=\sigma_{ji}

Hint: In eq. 1:

Conventional external power: 𝒲(Ωt)≔∫∂Ωtt⃗⋅v⃗+∫Ωtρb⃗⋅v⃗dv \mathcal{W}\left(\Omega_{t}\right)\coloneq{\int_{\partial\Omega_{t}}\vec{t}\cdot\vec{v}+\int_{\Omega_{t}}\rho\vec{b}\cdot\vec{v}dv}

Using Cauchy’s theorem and balance of linear momentum: ∫∂Ωtt⃗⋅v⃗da=∫∂Ωt(σ‾n⃗)⋅v⃗da=∫Ωt∇⋅(σ‾Tv⃗)dv=∫Ωt[v⃗⋅(∇⋅σ‾)+σ‾:∇v⃗]dv=∫Ωt[ρv⃗⋅v⃗̇−ρv⃗⋅b⃗+σ‾:D‾]dv=∫Ωt[ρv⃗⋅v⃗̇−ρv⃗⋅b⃗+σ‾:D‾]dv \begin{aligned} \int_{\partial\Omega_{t}}\vec{t}\cdot\vec{v}da&=\int_{\partial\Omega_{t}}\left(\bar{\sigma}\vec{n}\right)\cdot\vec{v}da\\ &=\int_{\Omega_{t}}\nabla\cdot\left(\bar{\sigma}^{T}\vec{v}\right)dv\\ &=\int_{\Omega_{t}}\left[\vec{v}\cdot\left(\nabla\cdot\bar{\sigma}\right)+\bar{\sigma}\colon\nabla{\vec{v}}\right]dv\\ &=\int_{\Omega_{t}}\left[\rho\vec{v}\cdot\dot{\vec{v}}-\rho\vec{v}\cdot\vec{b}+\bar{\sigma}\colon\bar{D}\right]dv\\ &=\int_{\Omega_{t}}\left[\rho\vec{v}\cdot\dot{\vec{v}}-\rho\vec{v}\cdot\vec{b}+\bar{\sigma}\colon\bar{D}\right]dv\\ \end{aligned}

Considering the above: ∫∂Ωtt⃗⋅v⃗+∫Ωtρb⃗⋅v⃗dv=∫Ωtρv⃗⋅v⃗̇dv+intΩtσ‾:D‾dv=∫Ωt12ρ|v⃗|2dv‾̇+∫Ωtσ‾:D‾dv \begin{aligned} \int_{\partial\Omega_{t}}\vec{t}\cdot\vec{v}+\int_{\Omega_{t}}\rho\vec{b}\cdot\vec{v}dv&=\int_{\Omega_{t}}\rho\vec{v}\cdot\dot{\vec{v}}dv+int_{\Omega_{t}}\bar{\sigma}\colon\bar{D}dv\\ &=\dot{\bar{\int_{\Omega_{t}}\dfrac{1}{2}\rho\lvert\vec{v}\rvert^{2}dv}}+\int_{\Omega_{t}}\bar{\sigma}\colon\bar{D}dv \end{aligned}

We have: 𝙴𝚡𝚝𝚎𝚛𝚗𝚊𝚕 𝚙𝚘𝚠𝚎𝚛=𝙺𝚒𝚗𝚎𝚝𝚒𝚌 𝚎𝚗𝚎𝚛𝚐𝚢+𝙸𝚗𝚝𝚎𝚛𝚗𝚊𝚕 𝚙𝚘𝚠𝚎𝚛 \texttt{External power} = \texttt{Kinetic energy} + \texttt{Internal power}

The external power is expended on the internal power and the change in kinetic energy.

2.5 The first law of thermodynamics

The external power expended on Ωt\Omega_{t} and net heat flow into Ωt\Omega_{t} results in increase of net internal energy and kinetic energy.

ℰ(Ωt)+𝒦(Ωt)‾̇=𝒲(Ωt)+𝒬(Ωt) \begin{aligned} \dot{\bar{\mathcal{E}\left(\Omega_{t}\right)+\mathcal{K}\left(\Omega_{t}\right)}} =\mathcal{W}\left(\Omega_{t}\right)+\mathcal{Q}\left(\Omega_{t}\right) \end{aligned}

We can write net internal energy and net heat flow as: ℰ(Ωt)=∫Ωtρϵdv𝒬(Ωt)=−∫∂Ωtq⃗⋅n⃗da+∫Ωtqdv \begin{aligned} \mathcal{E}\left(\Omega_{t}\right)&=\int_{\Omega_{t}}\rho\epsilon{dv}\\ \mathcal{Q}\left(\Omega_{t}\right)&=-\int_{\partial\Omega_{t}}\vec{q}\cdot\vec{n}da +\int_{\Omega_{t}}qdv\\ \end{aligned}

Using the mechanical power balance, we can write the first law of energy as: ∫Ωt(ρϵ̇−σ‾:D‾+∇⋅q⃗−q)dv=0ρϵ̇=σ‾:D‾−∇⋅q⃗+q \begin{aligned} \int_{\Omega_{t}}\left(\rho\dot{\epsilon}-\bar{\sigma}\colon\bar{D}+\nabla\cdot\vec{q}-q\right)dv=0\\ \rho\dot{\epsilon}=\bar{\sigma}\colon\bar{D}-\nabla\cdot\vec{q}+q \end{aligned}

2.6 The second law of thermodynamics

The net entropy production in Ωt\Omega_{t} is non-negative.

The rate of change of net internal-entropy 𝒮(Ωt)‾̇\dot{\bar{\mathcal{S}\left(\Omega_{t}\right)}} is sum of entropy flow 𝒥(Ωt)\mathcal{J}\left(\Omega_{t}\right) and net entropy production ℋ(Ωt)\mathcal{H}\left(\Omega_{t}\right) in Ωt\Omega_{t}.

ℋ(Ωt)≔𝒮(Ωt)‾̇−𝒥(Ωt) \begin{aligned} \mathcal{H}\left(\Omega_{t}\right)\coloneq\dot{\bar{\mathcal{S}\left(\Omega_{t}\right)}}-\mathcal{J}\left(\Omega_{t}\right) \end{aligned}

The second law of thermodynamics states: ℋ(Ωt)≥0 \begin{aligned} \mathcal{H}\left(\Omega_{t}\right)\geq{0} \end{aligned}

We can write the net internal entropy in terms of the specific entropy η\eta: 𝒮(Ωt)=∫Ωtρηdv \begin{aligned} \mathcal{S}\left(\Omega_{t}\right)=\int_{\Omega_{t}}\rho\eta{dv} \end{aligned}

and entropy flow in terms of entropy flux j⃗\vec{j} and entropy supply jj: 𝒥(Ωt)=−∫∂Ωtj⃗⋅n⃗+∫Ωtjdv \begin{aligned} \mathcal{J}\left(\Omega_{t}\right)=-\int_{\partial\Omega_{t}}\vec{j}\cdot\vec{n}+\int_{\Omega_{t}}jdv \end{aligned}

We can rewrite the second law of thermodynamics as: ∫Ωtρηdv‾̇≥−∫∂Ωtj⃗⋅n⃗da+∫Ωtjdv \begin{aligned} \dot{\bar{\int_{\Omega{t}}\rho\eta{dv}}}\geq-\int_{\partial\Omega{t}}\vec{j}\cdot\vec{n}da+\int_{\Omega_{t}}jdv\\ \end{aligned}

We have the following hypothesis: j⃗=q⃗θj=qθ \begin{aligned} \vec{j}=\dfrac{\vec{q}}{\theta}\\ j=\dfrac{q}{\theta} \end{aligned} Where θ\theta is the temperature.

Considering the above, we have ``Clausius-Duhem inequality" ∫Ωtρηdv‾̇≥−∫∂Ωtq⃗θ⋅n⃗da+∫Ωtqθdv \begin{aligned} \dot{\bar{\int_{\Omega_{t}}\rho\eta{dv}}}\geq-\int_{\partial\Omega{t}}\dfrac{\vec{q}}{\theta}\cdot\vec{n}da+\int_{\Omega_{t}}\dfrac{q}{\theta}dv\\ \end{aligned}

We can rewrite the entropy production as: ℋ(Ωt)=∫Ωt(ρη̇−qθ+∇⋅q⃗θ)dvℋ(Ωt)=∫ΩtΓdvΓ=ρη̇−qθ+∇⋅q⃗θ≥0 \begin{aligned} \mathcal{H}\left(\Omega_{t}\right)=\int_{\Omega_{t}}\left(\rho\dot{\eta}-\dfrac{q}{\theta}+\nabla\cdot\dfrac{\vec{q}}{\theta}\right)dv\\ \mathcal{H}\left(\Omega_{t}\right)=\int_{\Omega_{t}}\Gamma{dv}\\ \Gamma=\rho\dot{\eta}-\dfrac{q}{\theta}+\nabla\cdot\dfrac{\vec{q}}{\theta}\geq{0} \end{aligned}

Γ=ρη̇−qθ+1θ∇⋅q⃗−1θ2q⃗⋅∇θθΓ=ρθη̇−q+∇⋅q⃗−1θq⃗⋅∇θ−θΓ=q−∇⋅q⃗−ρθη̇+1θq⃗⋅∇θ−θΓ=ρė−σ‾:D‾−ρθη̇+1θq⃗⋅∇θ \begin{aligned} \Gamma&=\rho\dot{\eta}-\dfrac{q}{\theta}+\dfrac{1}{\theta}\nabla\cdot\vec{q}-\frac{1}{\theta^{2}}\vec{q}\cdot\nabla{\theta}\\ \theta\Gamma&=\rho\theta\dot{\eta}-q+\nabla\cdot\vec{q}-\frac{1}{\theta}\vec{q}\cdot\nabla{\theta}\\ -\theta\Gamma&=q-\nabla\cdot\vec{q}-\rho\theta\dot{\eta}+\frac{1}{\theta}\vec{q}\cdot\nabla{\theta}\\ -\theta\Gamma&=\rho\dot{e}-\bar{\sigma}\colon\bar{D}-\rho\theta\dot{\eta}+\frac{1}{\theta}\vec{q}\cdot\nabla{\theta}\\ \end{aligned}

ψ=e−θηψ̇=ė−θ̇η−θη̇ \begin{aligned} \psi&=e-\theta\eta\\ \dot{\psi}&=\dot{e}-\dot{\theta}\eta-\theta\dot{\eta} \end{aligned}

−∫ΩtθΓdv=∫Ωt(ρψ̇+ρηθ̇−σ‾:D‾+1θq⃗⋅∇θ)dv∫ΩtθΓdv=(∫∂Ωtt⃗⋅v⃗da+∫Ωtρb⃗⋅v⃗dv)−∫Ωtρ(ψ+12|v⃗|2)dv‾̇−∫Ωt(ρηθ̇+1θq⃗⋅∇θ)dv \begin{aligned} -\int_{\Omega_{t}}\theta\Gamma{dv}&=\int_{\Omega_{t}}\left(\rho\dot{\psi}+\rho\eta\dot{\theta}-\bar{\sigma}\colon\bar{D}+\dfrac{1}{\theta}\vec{q}\cdot\nabla\theta\right)dv\\ \int_{\Omega_{t}}\theta\Gamma{dv}&=\left(\int_{\partial\Omega_{t}}\vec{t}\cdot\vec{v}da+\int_{\Omega_{t}}\rho\vec{b}\cdot\vec{v}dv\right)-\dot{\bar{\int_{\Omega_{t}}\rho\left(\psi+\dfrac{1}{2}\lvert\vec{v}\rvert^{2}\right)dv}}-\int_{\Omega_{t}}\left(\rho\eta\dot{\theta}+\dfrac{1}{\theta}\vec{q}\cdot\nabla\theta\right)dv \end{aligned}

3 Strain

Stresses result in deformation; fig. 3 shows an initial configuration and corresponding deformed configuration.

Figure 3: Deformation

fig. 3 shows the initial and deformed configuration of a spatial region Ωt\Omega_{t}. The mapping x⃗=χ⃗(X⃗,t)\vec{x}=\vec{\chi}\left(\vec{X},t\right) denotes the motion of these configuration. At a fixed time, this mapping denotes the deformation. We have the following mapping for line element in the initial and deformed configuration.

dx⃗=∇X⃗χ⃗(X⃗,t)dX⃗=F‾(X⃗,t)dX⃗ \begin{aligned} d\vec{x}&=\nabla_{\vec{X}}\vec{\chi}\left(\vec{X},t\right)d\vec{X}\\ &=\bar{F}\left(\vec{X},t\right)d\vec{X} \end{aligned}

Mapping of the area element: n⃗da=dx1⃗×dx2⃗=F‾dX1⃗×F‾dX2⃗=𝚌𝚘𝚏(F‾)N⃗dA \begin{aligned} \vec{n}da&=d\vec{x_{1}}\times{d\vec{x_{2}}}\\ &=\bar{F}d\vec{X_{1}}\times{\bar{F}d\vec{X_{2}}}\\ &=\mathtt{cof}\left(\bar{F}\right)\vec{N}dA \end{aligned}

Mapping of the volume element: dv=𝚍𝚎𝚝(F‾)dV \begin{aligned} dv=\mathtt{det}\left(\bar{F}\right)dV \end{aligned}

Definition of strains: dx1⃗.dx2⃗−dX1⃗.dX2⃗=(F‾TF‾−I‾)dX1⃗.dX2⃗=2E‾dX1⃗.dX2⃗ \begin{aligned} d{\vec{x_{1}}}.d{\vec{x_{2}}}-d{\vec{X_{1}}}.d{\vec{X_{2}}}&= \left(\bar{F}^{T}\bar{F}-\bar{I}\right)d{\vec{X_{1}}}.d{\vec{X_{2}}}\\ &=2\bar{E}d{\vec{X_{1}}}.d{\vec{X_{2}}} \end{aligned}

Where E‾=12(F‾TF‾−I‾)\bar{E}=\dfrac{1}{2}\left(\bar{F}^{T}\bar{F}-\bar{I}\right) is the Green-Lagrangian strain tensor.

dx1⃗.dx2⃗−dX1⃗.dX2⃗=(I‾−F‾−TF‾−1)dx1⃗.dx2⃗=2Ê‾dx1⃗.dx2⃗ \begin{aligned} d{\vec{x_{1}}}.d{\vec{x_{2}}}-d{\vec{X_{1}}}.d{\vec{X_{2}}}&= \left(\bar{I}-\bar{F}^{-T}\bar{F}^{-1}\right)d{\vec{x_{1}}}.d{\vec{x_{2}}}\\ &=2\bar{\hat{E}}d{\vec{x_{1}}}.d{\vec{x_{2}}} \end{aligned}

Where Ê‾=12(I‾−F‾−TF‾−1)\bar{\hat{E}}=\dfrac{1}{2}\left(\bar{I}-\bar{F}^{-T}\bar{F}^{-1}\right) is the Eulerian-Almansi strain tensor.

The displacement field u(X⃗)u\left(\vec{X}\right) is: u⃗(X⃗)=x⃗−X⃗∇X⃗u⃗=F‾−I‾∇x⃗u⃗=I‾−F‾−1F‾=∇X⃗u⃗+I‾F‾−1=I‾−∇x⃗u⃗ \begin{aligned} \vec{u}\left(\vec{X}\right)&=\vec{x}-\vec{X}\\ \nabla_{\vec{X}}\vec{u}&=\bar{F}-\bar{I}\\ \nabla_{\vec{x}}\vec{u}&=\bar{I}-\bar{F}^{-1}\\ \bar{F}&=\nabla_{\vec{X}}\vec{u}+\bar{I}\\ \bar{F}^{-1}&=\bar{I}-\nabla_{\vec{x}}\vec{u} \end{aligned}

Now, the strain tensors can be written as: E‾=12(∇X⃗u⃗+(∇X⃗u⃗)T+(∇X⃗u⃗)T∇X⃗u⃗)Ê‾=12(∇x⃗u⃗+(∇x⃗u⃗)T−(∇x⃗u⃗)T∇x⃗u⃗) \begin{aligned} \bar{E}&=\dfrac{1}{2}\left(\nabla_{\vec{X}}\vec{u}+\left(\nabla_{\vec{X}}\vec{u}\right)^{T}+\left(\nabla_{\vec{X}}\vec{u}\right)^{T}\nabla_{\vec{X}}\vec{u}\right)\\ \bar{\hat{E}}&=\dfrac{1}{2}\left(\nabla_{\vec{x}}\vec{u}+\left(\nabla_{\vec{x}}\vec{u}\right)^{T}-\left(\nabla_{\vec{x}}\vec{u}\right)^{T}\nabla_{\vec{x}}\vec{u}\right) \end{aligned}

4 Coordinate transformation

The tensor is invariant to coordinate transform, however, the components changes with the basis. Tensor components transform with the following rule:

σ̃kl=QikσijQjl \begin{aligned} \tilde{\sigma}_{kl} &= Q_{ik}\sigma_{ij}Q_{jl} \end{aligned}

We derive this in the following: σ‾=σijei⊗ej \begin{aligned} \bar{\sigma} &= \sigma_{ij}e_{i}\otimes{e_{j}} \\ \end{aligned}

Coordinate transformation rule ẽi=Qikek \begin{aligned} \tilde{e}_{i} &= Q_{ik}e_{k} \\ \end{aligned}

Stress tensor remains invariant under coordinate transformation σ‾=σijQikQjlek⊗el=σ̃klek⊗el \begin{aligned} \bar{\sigma} &= \sigma_{ij}Q_{ik}Q_{jl}e_{k}\otimes{e_{l}}\\ &= \tilde{\sigma}_{kl}e_{k}\otimes{e_{l}}\\ \end{aligned}

By comparing the components, we have: σ̃kl=QikσijQjl \begin{aligned} \tilde{\sigma}_{kl} &= Q_{ik}\sigma_{ij}Q_{jl} \end{aligned}

4.1 Three dimensional stress tensor

For a plane passing through the point with stress tensor as σij\sigma_{ij},

We evaluate the traction as: t⃗=σ‾n⃗ \vec{t}=\bar{\sigma}\vec{n}

Correspondingly, the normal and shear stress are: σn=n⃗⋅t⃗=n⃗⋅σ‾n⃗=σ1n12+σ2n22+σ3n32τn=|t⃗|2−σn2σn2+τn2=σ12n12+σ22n22+σ32n32 \begin{aligned} \sigma_{n}&=\vec{n}\cdot\vec{t}\\ &=\vec{n}\cdot\bar{\sigma}\vec{n}\\ &=\sigma_{1}n_{1}^{2}+\sigma_{2}n_{2}^{2}+\sigma_{3}n_{3}^{2}\\ \tau_{n}&=\sqrt{\lvert\vec{t}\rvert^{2}-\sigma_{n}^{2}}\\ \sigma_{n}^{2}+\tau_{n}^{2}&=\sigma_{1}^{2}n_{1}^{2}+\sigma_{2}^{2}n_{2}^{2}+\sigma_{3}^{2}n_{3}^{2}\\ \end{aligned}

With the above equation for τn\tau_{n}, we have two signs.

We also have n12+n22+n33=1 n_{1}^{2}+n_{2}^{2}+n_{3}^{3}=1

We have the set of linear equations: [111σ1σ2σ3σ12σ22σ32]{n12n22n32}={1σnσn2+τn2} \begin{aligned} \begin{bmatrix} 1 & 1 & 1 \\ \sigma_{1} & \sigma_{2} & \sigma_{3}\\ \sigma_{1}^{2} & \sigma_{2}^{2} & \sigma_{3}^{2} \end{bmatrix} \begin{Bmatrix} n_{1}^{2}\\ n_{2}^{2}\\ n_{3}^{2} \end{Bmatrix} &= \begin{Bmatrix} 1\\ \sigma_{n}\\ \sigma_{n}^{2}+\tau_{n}^{2} \end{Bmatrix} \end{aligned}

Assuming σ1≠σ2≠σ3\sigma_{1}\neq\sigma_{2}\neq\sigma_{3}:

{n12n22n32}=[σ2σ3(σ1−σ2)(σ1−σ3)−σ2+σ3(σ1−σ2)(σ1−σ3)1(σ1−σ2)(σ1−σ3)σ3σ1(σ2−σ3)(σ2−σ1)−σ3+σ1(σ2−σ3)(σ2−σ1)1(σ2−σ3)(σ2−σ1)σ1σ2(σ3−σ1)(σ3−σ2)−σ1+σ2(σ3−σ1)(σ3−σ2)1(σ3−σ1)(σ3−σ2)]{1σnσn2+τn2} \begin{aligned} \begin{Bmatrix} n_{1}^{2}\\ n_{2}^{2}\\ n_{3}^{2} \end{Bmatrix} &= \begin{bmatrix} \frac{\sigma_{2}\sigma_{3}}{\left(\sigma_{1}-\sigma_{2}\right)\left(\sigma_{1}-\sigma_{3}\right)} & -\frac{\sigma_{2}+\sigma{3}}{\left(\sigma_{1}-\sigma_{2}\right)\left(\sigma_{1}-\sigma_{3}\right)} & \frac{1}{\left(\sigma_{1}-\sigma_{2}\right)\left(\sigma_{1}-\sigma_{3}\right)}\\ \frac{\sigma_{3}\sigma_{1}}{\left(\sigma_{2}-\sigma_{3}\right)\left(\sigma_{2}-\sigma_{1}\right)} & -\frac{\sigma_{3}+\sigma{1}}{\left(\sigma_{2}-\sigma_{3}\right)\left(\sigma_{2}-\sigma_{1}\right)} & \frac{1}{\left(\sigma_{2}-\sigma_{3}\right)\left(\sigma_{2}-\sigma_{1}\right)}\\ \frac{\sigma_{1}\sigma_{2}}{\left(\sigma_{3}-\sigma_{1}\right)\left(\sigma_{3}-\sigma_{2}\right)} & -\frac{\sigma_{1}+\sigma{2}}{\left(\sigma_{3}-\sigma_{1}\right)\left(\sigma_{3}-\sigma_{2}\right)} & \frac{1}{\left(\sigma_{3}-\sigma_{1}\right)\left(\sigma_{3}-\sigma_{2}\right)}\\ \end{bmatrix} \begin{Bmatrix} 1\\ \sigma_{n}\\ \sigma_{n}^{2}+\tau_{n}^{2} \end{Bmatrix} \end{aligned}

With n12≥0,n22≥0,n32≥0n_{1}^{2}\geq{0},n_{2}^{2}\geq{0},n_{3}^{2}\geq{0}, and assuming σ1≥σ2≥σ3\sigma_{1}\geq\sigma_{2}\geq\sigma_{3}:

[σn−(σ2+σ32)]2+τn2≤(σ2−σ32)2][σn−(σ3+σ12)]2+τn2≥(σ3−σ22)2][σn−(σ1+σ22)]2+τn2≤(σ1−σ22)2] \begin{aligned} \left[\sigma_{n}-\left(\dfrac{\sigma_{2}+\sigma_{3}}{2})\right]^{2}+\tau_{n}^{2}\leq\left(\dfrac{\sigma_{2}-\sigma_{3}}{2}\right)^{2}\right]\\ \left[\sigma_{n}-\left(\dfrac{\sigma_{3}+\sigma_{1}}{2})\right]^{2}+\tau_{n}^{2}\geq\left(\dfrac{\sigma_{3}-\sigma_{2}}{2}\right)^{2}\right]\\ \left[\sigma_{n}-\left(\dfrac{\sigma_{1}+\sigma_{2}}{2})\right]^{2}+\tau_{n}^{2}\leq\left(\dfrac{\sigma_{1}-\sigma_{2}}{2}\right)^{2}\right]\\ \end{aligned}

Figure 4: Three dimensional Mohr circle

fig. 4 shows the admissible stress states.

4.2 Two-dimensional stress tensor

fig. 5 shows the original coordinate system x1−x2x_{1}-x_{2} and transformed coordinate system x̃1−x̃2\tilde{x}_{1}-\tilde{x}_{2}. The rotation matrix is given by:

Figure 5: Coordinate transformation in two-dimension

Qij=ẽi⋅ej[Qij]=[cosθsinθ−sinθcosθ] \begin{aligned} Q_{ij} &= \tilde{e}_{i}\cdot{e}_{j}\\ \left[Q_{ij}\right] &= \begin{bmatrix} \cos{\theta} & \sin{\theta}\\ -\sin{\theta} & \cos{\theta} \end{bmatrix} \end{aligned}

The stress components in the transformed coordinate system: σ̃ij=QkiσklQljσ̃11=σ11+σ222+σ11−σ222cos2θ+σ12sin2θσ̃12=σ12cos2θ+σ22−σ112sin2θ \begin{aligned} \tilde{\sigma}_{ij} &= Q_{ki}\sigma_{kl}Q_{lj}\\ \tilde{\sigma}_{11} &= \dfrac{\sigma_{11}+\sigma_{22}}{2}+\dfrac{\sigma_{11}-\sigma_{22}}{2}\cos{2\theta}+\sigma_{12}\sin{2\theta}\\ \tilde{\sigma}_{12} &= \sigma_{12}\cos{2\theta}+\dfrac{\sigma_{22}-\sigma_{11}}{2}\sin{2\theta} \end{aligned}

If we desire to estimate the normal and shear stress on the plane normal to x̃1\tilde{x}_{1}: σn−σ11+σ222=σ11−σ222cos2θ+σ12sin2θ=Rsin(2θ+ϕ)τn=σ12cos2θ−σ11−σ222sin2θ=Rcos(2θ+ϕ)R=σ122+(σ11−σ222)2tanϕ=σ11−σ222σ12 \begin{aligned} \sigma_{n}-\dfrac{\sigma_{11}+\sigma_{22}}{2} &= \dfrac{\sigma_{11}-\sigma_{22}}{2}\cos{2\theta}+\sigma_{12}\sin{2\theta}\\ &= R\sin\left(2\theta+\phi\right)\\ \tau_{n} &= \sigma_{12}\cos{2\theta}-\dfrac{\sigma_{11}-\sigma_{22}}{2}\sin{2\theta}\\ &= R\cos\left(2\theta+\phi\right)\\ R &= \sqrt{\sigma_{12}^{2}+\left(\dfrac{\sigma_{11}-\sigma_{22}}{2}\right)^{2}}\\ \tan\phi &= \dfrac{\sigma_{11}-\sigma_{22}}{2\sigma_{12}} \end{aligned}

The locus of normal and shear stress is a circle, called Mohr circle.

5 Mohr Circle

The Mohr circle is very useful in quickly extracting the tensor values on different planes or in extracting the planes corresponding to different tensor values. The method of poles in particularly interesting in Mohr circles. It only works with the right sign conventions:

In the following, we see how pole appears with compression positive and counterclockwise shear positive.

Figure 6: Stress-state

For an element with the stress state given in the fig. 6, we have normal stress and shear sterss on the plane normal to n⃗=cosθe⃗1+sinθe⃗2\vec{n}=\cos{\theta}\vec{e}_{1}+\sin{\theta}\vec{e}_{2} as:

σn=σ11+σ222+σ11−σ222cos2θ+σ12sin2θ=σ11+σ222+Rsin(2θ+ϕ)τn=σ11−σ222sin2θ−σ12cos2θ=−Rcos(2θ+ϕ)R=(σ11−σ222)2+σ122tanϕ=σ11−σ222σ12 \begin{aligned} \sigma_{n} &= \dfrac{\sigma_{11}+\sigma_{22}}{2}+\dfrac{\sigma_{11}-\sigma_{22}}{2}\cos{2\theta}+\sigma_{12}\sin{2\theta}\\ &= \dfrac{\sigma_{11}+\sigma_{22}}{2}+R\sin\left(2\theta+\phi\right)\\ \tau_{n} &= \dfrac{\sigma_{11}-\sigma_{22}}{2}\sin{2\theta}-\sigma_{12}\cos{2\theta}\\ &= -R\cos\left(2\theta+\phi\right)\\ R &= \sqrt{\left(\dfrac{\sigma_{11}-\sigma_{22}}{2}\right)^{2}+\sigma_{12}^{2}}\\ \tan{\phi} &= \dfrac{\sigma_{11}-\sigma_{22}}{2\sigma_{12}}\\ \end{aligned} 

# import the libraries

import numpy as np
from matplotlib import pyplot as plt

# stress state at the point
s11 = 150
s12 = 50
s22 = 50

# radius and center of the Mohr circle
R = np.sqrt(((s11-s22)/2)**2+s12**2)
center = (s11+s22)/2
phi = np.arctan((s11-s22)/(2*s12)) # twice the inclination of principal plane

# number of points on the Mohr circle and lenght of the plane
num_points = 100
length_plane = 400

# stress states on the Mohr circle for different theta values
snfn = lambda t: center + R * np.sin(2*t+phi)
tnfn = lambda t: -1 * R * np.cos(2*t+phi)
theta = np.linspace(0, 2*np.pi, num=num_points)
sn = snfn(theta)
tn = tnfn(theta)

'''
plot the Mohr circle and the planes on which stresses, at the point in the
Mohr circle, acts.
From [@fig:StressState] direction cosines of the plane are
[cos(270+theta), sin(270+theta)] --> [sin(theta), -cos(theta)]
The line that represents plane on the Mohr circle has start point
$\left(x-\frac{l}{2}\sin{\theta},y+\frac{l}{2}\cos{\theta}\right)$ and end
point $\left(x+\frac{l}{2}\sin{\theta},y-\frac{l}{2}\cos{\theta}\right)$. The
point $\left(x,y\right)$ is the point on the Mohr circle.
'''
plt.figure()
plt.plot(sn, tn)
for i in np.linspace(0, 2*np.pi, num =10):
    # point on the Mohr circle
    plt.plot(snfn(i), tnfn(i), 'o', MarkerSize=20)
    # start and end points
    start = [snfn(i)-length_plane/2*np.sin(i),tnfn(i)+length_plane/2*np.cos(i)]
    end = [snfn(i)+length_plane/2*np.sin(i),tnfn(i)-length_plane/2*np.cos(i)]
    point_x = [start[0], end[0]]
    point_y = [start[1], end[1]]
    plt.plot(point_x, point_y, '-')
plt.axis('equal')
plt.xlabel(r'Normal stress $\sigma_{n}$')
plt.ylabel(r'Shear stress $\tau_{n}$')
Pole

5.1 Mohr circle of strain

fig. 7 shows the strain state in an element. We consider contraction and counterclockwise shear as positive. In the fig. 7, strain state is the following:

dϵ11=δl11l1dϵ12=dγ122=12(δl12l1+δl21l2)dϵ21=dγ212=−12(δl12l1+δl21l2)dϵ22=−δl22l2 \begin{aligned} d\epsilon_{11} &= \dfrac{\delta{l}_{11}}{l_{1}}\\ d\epsilon_{12} &= \dfrac{d\gamma_{12}}{2}\\ &= \dfrac{1}{2}\left(\dfrac{\delta{l}_{12}}{l_{1}}+\dfrac{\delta{l}_{21}}{l_{2}}\right)\\ d\epsilon_{21} &= \dfrac{d\gamma_{21}}{2}\\ &= -\dfrac{1}{2}\left(\dfrac{\delta{l}_{12}}{l_{1}}+\dfrac{\delta{l}_{21}}{l_{2}}\right)\\ d\epsilon_{22} &= -\dfrac{\delta{l}_{22}}{l_{2}} \end{aligned}

Note that the sign convention here is according the Mohr circle and the above does not imple the strain tensor is assymetric. The engineering shear strain dγ12d\gamma_{12} is the change in angle between two perpendicular line elements in the initial configuration.

Figure 7: Strain state

fig. 8 shows the Mohr circle corrsponding to this strain state. The strain components dϵ11,dϵ12d\epsilon_{11}, d\epsilon_{12} act on the vertical plane; we obtain the pole by drawing the vertical line at dϵ11,dϵ12d\epsilon_{11}, d\epsilon_{12} which intersects the Mohr circle at the pole. The line joining the pole and zero normal strain points correspond to plane of zero extension and line perpendicular to that is the zero extension line.

Figure 8: Strain Mohr Circle

5.2 Dilatancy

Granular media changes its volume upon distortion; this property is known as dilatancy.

Figure 9: Dilatancy

fig. 9 shows an element under distortion along x1x_{1}; there is no contraction or dilation along this direction. The line of zero extension lies along x1x_{1}. Angle of dilatancy (ψ\psi) is defined as

tanψ=δl22δl21=−dϵ22dγz=−dϵvdγz \begin{aligned} \tan\psi &= \dfrac{\delta{l}_{22}}{\delta{l}_{21}}\\ &= -\dfrac{d\epsilon_{22}}{d\gamma_{z}}\\ &= -\dfrac{d\epsilon_{v}}{d\gamma_{z}} \end{aligned}

Slip surfaces in soil mass have the shape of logarithmic spiral. When a contiguous elements reach the failure in soil, a slip surface forms. The soil around the slip surface behaves as a rigid block such that distortion occurs within the slip surface. The tangents to slip surfaces are lines of zero extension. From the fig. 10:

tanψ=rdθdrlogr|r=r0r=θtanψ|]θ=0θr=r0exp(θtanψ) \begin{aligned} \tan{\psi} &= \dfrac{rd\theta}{dr}\\ \log{r}\rvert_{r=r_{0}}^{r} &= \theta\tan{\psi}\rvert]_{\theta=0}^{\theta}\\ r &= r_{0}\exp\left(\theta\tan\psi\right) \end{aligned}

Figure 10: Slip surface

6 Failure criteria

6.1 Mohr-Coulomb failure criteria

τ=c+σtanϕt=ccosϕ+ssinϕσ1=σ3Nϕ+2cNϕ \begin{aligned} \tau &= c+\sigma\tan{\phi}\\ t &= c\cos{\phi}+s\sin{\phi}\\ \sigma_{1} &= \sigma_{3}N_{\phi}+2c\sqrt{N{\phi}} \end{aligned}

Where σ1\sigma_{1} and σ3\sigma_{3} are major and minor principal stresses, and s=σ1+σ32s=\frac{\sigma_{1}+\sigma_{3}}{2}, t=σ1−σ32t=\frac{\sigma_{1}-\sigma_{3}}{2} are the peak of the Mohr-Circle.

6.2 Hoek-Brown failure criteria

σ1=σ3+qumσ3qu+s \begin{aligned} \sigma_{1} = \sigma_{3}+q_{u}\sqrt{m\frac{\sigma_{3}}{q_{u}}+s} \end{aligned}

7 Stress discontinuity: rotation of principal stress

In a boundary value problem, the principal planes of stresses or strains rotate.

7.1 Finite rotation over a stress discontinuity

7.1.1 Cohesive soil under undrained loading

7.1.1.1 simple case

Figure 11: Stress rotation behind a frictional retaining wall

7.1.1.2 General case

We have a stress discontinuity over a plane which makes angle xx, in counterclockwise sense, with the vertical. The principal planes are neither horizontal or verical. The Mohr cirlce on both sides of disconituity have the following properties fig. 12.

Figure 12: Stress rotation under general condition
Mean stress Pole Plane of discontinuity Principal plane rotation
s0s_{0} yy xx θ=π/2+2x−y\theta=\pi/2+2x-y
s1s_{1} y+2θy+2\theta –

s1=s0+2cusin(θ) \begin{aligned} s_{1}=s_{0}+2c_{u}\sin\left(\theta\right) \end{aligned}

7.1.2 Cohesion-less soil

Figure 13: Stress rotation in frictional soil

7.2 Infinitesimal rotation over a stress discontinuity

A soil mass fails when a contiguous failure surface forms such that stress state at each point satisfies Mohr-Coulomb failure criteria. An example stress rotation is given in fig. 14.

Figure 14: Stress rotation

For zone A, we can write the following –

σn=σ1A+σ3A2+σ1A−σ3A2cos2θτn=σ1A−σ3A2sin2θ \begin{aligned} \sigma_{n} &= \dfrac{\sigma_{1A}+\sigma_{3A}}{2} + \dfrac{\sigma_{1A}-\sigma_{3A}}{2}\cos{2\theta}\\ \tau_{n} &= \dfrac{\sigma_{1A}-\sigma_{3A}}{2}\sin{2\theta} \end{aligned}

For zone B, we can write the following –

σn=σ1B+σ3B2+σ1B−σ3B2cos(2θ+2δθ)τn=σ1B−σ3B2sin(2θ+2δθ) \begin{aligned} \sigma_{n} &= \dfrac{\sigma_{1B}+\sigma_{3B}}{2} + \dfrac{\sigma_{1B}-\sigma_{3B}}{2}\cos\left(2\theta+2\delta\theta\right)\\ \tau_{n} &= \dfrac{\sigma_{1B}-\sigma_{3B}}{2}\sin\left(2\theta+2\delta\theta\right) \end{aligned}

In general, the shear and normal stress at the discontinuity can be written as –

σn=s+tcos2θτn=tsin2θ \begin{aligned} \sigma_{n} &= s + t\cos{2\theta}\\ \tau_{n} &= t\sin{2\theta} \end{aligned}

At the zone of discontinuity, we also have –

dσn=0dτn=0 \begin{aligned} d\sigma_{n} &= 0 \\ d\tau_{n} &= 0 \\ \end{aligned}

This implies –

ds+dtcos2θ−2tsin2θδθ=0dtsin2θ+2tcos2θδθ=0 \begin{aligned} ds + dt\cos{2\theta} - 2t\sin{2\theta}\delta\theta &=0 \\ dt\sin{2\theta} + 2t\cos{2\theta}\delta\theta &= 0\\ \end{aligned}

7.2.1 Cohesive soils

For the cohesive soil, the Mohr-Coulomb failure criteria is given as:

τ=cu \begin{aligned} \tau = c_{u} \end{aligned}

Where cuc_{u} is the undrained cohesive strength.

The radius of Mohr circle doesn’t change at failure –

ds=2cuδθ \begin{aligned} ds = 2c_{u}\delta\theta \end{aligned}

Note 2θ=π22\theta=\frac{\pi}{2}.

7.2.2 Cohesion-less soils

For a cohesion-less soil, the Mohr-Coulomb failure criteria is given as:

τ=σtanϕt=ssinϕ \begin{aligned} \tau &= \sigma\tan{\phi} \\ t &= s\sin{\phi} \end{aligned}

Also 2θ=π2+ϕ2\theta=\frac{\pi}{2}+\phi.

ds−dtsinϕ−2tcosϕδθ=0dtcosϕ−2tsinϕδθ=0 \begin{aligned} ds - dt\sin{\phi} - 2t\cos{\phi}\delta\theta &=0 \\ dt\cos{\phi} - 2t\sin{\phi}\delta\theta &= 0\\ \end{aligned}

From the above, we have –

ds=2tδθ(cosϕ+sin2ϕcosϕ)ds=2s(tanϕ)δθ \begin{aligned} ds &= 2t\delta\theta\left(\cos\phi+\frac{\sin^{2}\phi}{\cos{\phi}}\right) \\ ds &= 2s\left(\tan{\phi}\right)\delta\theta \end{aligned}

8 Earth retaining structures

At rest,

ϵh=01E[σh−νσv−νσh]=0K0=σhσv=ν1−ν \begin{aligned} \epsilon_{h} &= 0\\ \dfrac{1}{E}\left[\sigma_{h}-\nu\sigma_{v}-\nu\sigma_{h}\right] &= 0\\ K_{0} &= \dfrac{\sigma_{h}}{\sigma_{v}}\\ &= \dfrac{\nu}{1-\nu} \end{aligned}

8.1 Frictionless interaction between soil and structure

8.1.1 Rankine theory

8.1.1.1 Flat surface

Figure 15: Active and passive earth pressure for leveled soil mass behind a vertical structure

8.1.1.2 Inclined surface

Figure 16: Active and passive earth pressure for inclined soil mass behind a vertical structure

8.2 Frictional interaction between soil and structure

8.2.1 Coulomb theory

8.2.2 Stress characteristics

9 Readings

A good knowledge of continuum mechanics helps in understanding soil mechanics and soil plasticity.

  1. Gurtin, M.E., Fried, E. and Anand, L., 2010. The mechanics and thermodynamics of continua. Cambridge University Press.
  2. http://solidmechanics.org/contents.php [highly recommended]
  3. Jog, C.S., 2015. Continuum mechanics (Vol. 1). Cambridge University Press. [My teacher’s book.]